The series given is $2 + 12 + 36 + 80 + \dots$. To find the sum of the first 20 terms, we need to identify the pattern in the series.

Step 1: Identify the pattern

Let's observe the pattern in the sequence:

• $12 - 2 = 10$
• $36 - 12 = 24$
• $80 - 36 = 44$

The differences between consecutive terms (10, 24, 44) suggest that this is not an arithmetic sequence. Instead, the differences themselves seem to follow a pattern. Let's check the second differences:

• $24 - 10 = 14$
• $44 - 24 = 20$

Since the second differences are constant, this suggests that the sequence might be a quadratic sequence.

Step 2: Assume a quadratic formula

Assume the general term of the sequence $T_n$ is given by: $T_n = an^2 + bn + c$

Using the first three terms to form equations:

• $T_1 = 2$: $a(1)^2 + b(1) + c = 2$ $\rightarrow a + b + c = 2$
• $T_2 = 12$: $a(2)^2 + b(2) + c = 12$ $\rightarrow 4a + 2b + c = 12$
• $T_3 = 36$: $a(3)^2 + b(3) + c = 36$ $\rightarrow 9a + 3b + c = 36$

We now have the system of equations:

1. $a + b + c = 2$
2. $4a + 2b + c = 12$
3. $9a + 3b + c = 36$

Step 3: Solve the system of equations

Subtract equation 1 from equation 2: $(4a + 2b + c) - (a + b + c) = 12 - 2$ $3a + b = 10 \quad \text{(Equation 4)}$

Subtract equation 2 from equation 3: $(9a + 3b + c) - (4a + 2b + c) = 36 - 12$ $5a + b = 24 \quad \text{(Equation 5)}$

Subtract equation 4 from equation 5: $(5a + b) - (3a + b) = 24 - 10$ $2a = 14 \quad \Rightarrow a = 7$

Substitute $a = 7$ into Equation 4: $3(7) + b = 10 \quad \Rightarrow 21 + b = 10 \quad \Rightarrow b = -11$

Substitute $a = 7$ and $b = -11$ into Equation 1: $7 - 11 + c = 2 \quad \Rightarrow -4 + c = 2 \quad \Rightarrow c = 6$

Thus, the general term $T_n$ is: $T_n = 7n^2 - 11n + 6$

Step 4: Find the sum of the first 20 terms

The sum of the first $n$ terms of the sequence is: $S_n = \sum_{n=1}^{20} T_n = \sum_{n=1}^{20} (7n^2 - 11n + 6)$

The sum of the squares, linear terms, and constants are: $S_n = 7\sum_{n=1}^{20} n^2 - 11\sum_{n=1}^{20} n + 6\sum_{n=1}^{20} 1$

Using the sum formulas: $\sum_{n=1}^{20} n^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{n=1}^{20} n = \frac{n(n+1)}{2}, \quad \sum_{n=1}^{20} 1 = 20$ Substitute $n = 20$: $\sum_{n=1}^{20} n^2 = \frac{20(21)(41)}{6} = 2870$ $\sum_{n=1}^{20} n = \frac{20(21)}{2} = 210$ $\sum_{n=1}^{20} 1 = 20$

Thus: $S_{20} = 7(2870) - 11(210) + 6(20) = 20090 - 2310 + 120 = 17900$

So, the sum of the first 20 terms is $17900$.

If you need more details or have any questions, feel free to ask! Here are five related questions:

1. How would the sum change if we extended the series to 30 terms?
2. What if the sequence was geometric instead of quadratic? How would that change the sum calculation?
3. Can you derive the sum formula for an arithmetic sequence?
4. What would happen to the sum if the sequence started at a different term?
5. How can you verify that the derived formula for $T_n$ is correct?

Tip: When identifying the pattern in a sequence, always check for differences between terms; if the first differences are not constant, check the second differences to see if the sequence is quadratic.